Question: The scalar field $f(x, y) = x^3 + y^3 - x - y$ has a critical point at $\left( \dfrac{\sqrt{3}}{3}, \dfrac{\sqrt{3}}{3} \right)$. How does the second partial derivative test classify this critical point? Choose 1 answer: Choose 1 answer: (Choice A) A Local maximum (Choice B) B Local minimum (Choice C) C Saddle point (Choice D, Checked) D The test is inconclusive
Solution: The second partial derivative test uses the quantity below, evaluated at the critical point we wish to classify. $H = f_{xx}f_{yy} - f_{xy}f_{yx}$ $H < 0$ implies a saddle point. $H > 0$ and $f_{xx} > 0$ implies a local minimum. $H > 0$ and $f_{xx} < 0$ implies a local minimum. $H = 0$ means the test is inconclusive. Let's calculate $H$. First we need all the regular partial derivatives. $\begin{aligned} f_x &= 3x^2 - 1 \\ \\ f_y &= 3y^2 - 1 \end{aligned}$ Now we can find all the second order partial derivatives. $\begin{aligned} f_{xx} &= 6x = 2\sqrt{3} \\ \\ f_{yx} &= 0 \\ \\ f_{xy} &= 0 \\ \\ f_{yy} &= 6y = 2\sqrt{3} \end{aligned}$ Therefore, $H = (2\sqrt{3})(2\sqrt{3}) - (0)(0) = 12$. Because $H$ is positive, the critical point is either a local maximum or a local minimum. To find which, we can notice that $f_{xx} > 0$, so the critical point is a local minimum.